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Question: If a raw egg and a boiled egg are spin together with the same angular velocity on the horizontal sur...

If a raw egg and a boiled egg are spin together with the same angular velocity on the horizontal surface then which one will stop first?
A) Boiled egg
B) Raw egg
C) Both egg will stop simultaneously
D) Cannot be ascertained.

Explanation

Solution

Estimations are introduced on the ascent of a turning egg. It was discovered that the turn, the angular momentum and the active kinetic energy all abate as the egg rises, dissimilar to the instance of a ballet dancer who can build her turn and active energy by lessening her snapshot of inactivity. The observed impacts can be clarified to a limited extent, as far as moving contact between the egg and the surface on which it turns.

Formula used: When a torque applied to a body, it starts rotating about its axis with angular acceleration. The relation of torque and angular acceleration is –
τ=Iα\tau = I\alpha
Where IIis the moment of inertia of the body and ω\omega be the angular velocity of the body, jjbe the angular momentum then, moment of inertia is
j=I×ω\overline j = I \times \overline \omega

Complete step by step solution: -
We have two eggs. One is raw and the second is boiled. Raw egg has liquid inside it. Hence it is considered as a rigid body with fluid inside it. ON the other hand, the boiled egg has a hard substance inside it. So, it is considered as a rigid body.
We know that the moment of a rigid body is always greater than the fluid body.
If Ir{I_r}and Ib{I_b}be the moment of inertia of raw egg and boiled egg respectively. So,
Ir<Ib{I_r} < {I_b}........................(i)
And by the relation of the torque and angular acceleration, we have-
τ=Iα\tau = I\alpha
If τ\tau is same for both the eggs (condition according to the question)
So,
I1αI \propto \dfrac{1}{\alpha }
Now, moment of inertia for raw egg-
Ir1αr{I_r} \propto \dfrac{1}{{{\alpha _r}}}........................(ii)
And moment of inertia for boiled egg-
Ib1αb{I_b} \propto \dfrac{1}{{{\alpha _b}}}............................(iii)
Now, we are finding the time, in which both eggs will stop. So, we can justify which egg will stop first.
When raw egg will stop, its angular velocity becomes zero i.e. ω=0{\omega '} = 0
Now,
ω0=ω{\omega _0} = \omega (Same for both eggs)
α=αr\alpha = - \alpha {}_r (Because αr{\alpha _r}is the retardation)
t=trt = {t_r}
Now, from first equation of motion-
ω=ω0+αt 0=ωαrtr ω=αrtr  {\omega '} = {\omega _0} + \alpha t \\\ \Rightarrow 0 = \omega - {\alpha _r}{t_r} \\\ \Rightarrow \omega = {\alpha _r}{t_r} \\\
tr=ωαr\Rightarrow {t_r} = \dfrac{\omega }{{{\alpha _r}}} ……………..(iv)
Similarly, for boiled egg, the quantities below are the same as raw egg.
ω=0 ω0=ω α=αb t=tb  {\omega ^{''}} = 0 \\\ {\omega _0} = \omega \\\ \alpha = - {\alpha _b} \\\ t = {t_b} \\\
Again by first equation of motion,
ω=ω0+αt 0=ωαbtb ω=αbtb  {\omega ^{''}} = {\omega _0} + \alpha t \\\ \Rightarrow 0 = \omega - {\alpha _b}{t_b} \\\ \Rightarrow \omega = {\alpha _b}{t_b} \\\
tb=ωαb\Rightarrow {t_b} = \dfrac{\omega }{{{\alpha _b}}}...................(v)
Dividing equation (iv) by equation (v), we get-
trtb=ωαrωαb\dfrac{{{t_r}}}{{{t_b}}} = \dfrac{{\dfrac{\omega }{{{\alpha _r}}}}}{{\dfrac{\omega }{{{\alpha _b}}}}}
trtb=αbαr\dfrac{{{t_r}}}{{{t_b}}} = \dfrac{{{\alpha _b}}}{{{\alpha _r}}}.....................(vi)
Now, from equation (ii) and (iii), we get-
IrIb=αbαr\dfrac{{{I_r}}}{{{I_b}}} = \dfrac{{{\alpha _b}}}{{{\alpha _r}}}
Now, from equation (vi), we get the following equation after substituting the values.
trtb=IrIb\dfrac{{{t_r}}}{{{t_b}}} = \dfrac{{{I_r}}}{{{I_b}}}
We discussed earlier thatIr<Ib{I_r} < {I_b}. So, we get
tr<tb{t_r} < {t_b}
Hence, raw egg will stop first.

Therefore, option B is correct.

Note: - We have to remember that the moment of inertia is directly proportional to the time taken by the body and all the rotational measurements are related to the moment of inertia.